Chapter 11, Problem 6P

(a)

To determine

The value of ${\cos }^{-1}\left[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right]$.

Answer to Problem 6P

The value of given expression ${\cos }^{-1}\left[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right]$ is $\underset{\_}{168°}$.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{A}$ the vector as.

  $A=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}$                                                                           (I)

Here, $A_x$ is $x$-component of $\overrightarrow{A}$, $A_y$ is $y$-component of $\overrightarrow{A}$, $A_z$ is $z$-component of $\overrightarrow{A}$ and $A$ is the magnitude of $\overrightarrow{A}$.

Write the expression for the magnitude of $\overrightarrow{B}$ the vector as.

  $B=\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}$                                                                         (II)

Here, $B_x$ is $x$-component of $\overrightarrow{B}$, $B_y$ is $y$-component of $\overrightarrow{B}$, $B_z$ is $z$-component of $\overrightarrow{B}$ and $B$ is the magnitude of $\overrightarrow{B}$.

Write the expression for dot product of a vector $\overrightarrow{A}$ and $\overrightarrow{B}$ as.

  $\left(\overrightarrow{A}\cdot \overrightarrow{B}\right)=\left(A_x\cdot B_x\right)+\left(A_y\cdot B_y\right)+\left(A_z\cdot B_z\right)$                                                    (III)

Consider the angle for a given expression is $\theta$.

Write the expression for the angle between the vectors.

  $\theta ={\cos }^{-1}\left[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right]$

Substitute $\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}$ for $A$, $\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}$ for $B$ and $\left(A_x\cdot B_x\right)+\left(A_y\cdot B_y\right)+\left(A_z\cdot B_z\right)$ for $\left(\overrightarrow{A}\cdot \overrightarrow{B}\right)$ in the above expression.

  $\theta ={\cos }^{-1}\left[\frac{\left(\left(A_x\cdot B_x\right)+\left(A_y\cdot B_y\right)+\left(A_z\cdot B_z\right)\right)}{\left(\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}\right)\left(\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}\right)}\right]$             (IV)

Conclusion:

Substitute $-3$ for $A_x$, $7$ for $A_y$, $-4$ for $A_z$, $6$ for $B_x$, $-10$ for $B_y$ and $9$ for $B_z$ in equation (VI).

  $\begin{array}{c}\theta ={\cos }^{-1}\left[\frac{\left(\left(\left(-3\right)6\right)+\left(7\left(-10\right)\right)+\left(\left(-4\right)9\right)\right)}{\left(\sqrt{{\left(-3\right)}^2+{\left(7\right)}^2+{\left(-4\right)}^2}\right)\left(\sqrt{{\left(6\right)}^2+{\left(-10\right)}^2+{\left(9\right)}^2}\right)}\right]\\ ={\cos }^{-1}\left[\frac{\left(-124\right)}{\left(\sqrt{74}\right)\left(\sqrt{217}\right)}\right]\\ =167.52°\\ \approx 168°\end{array}$

Thus, the value of the given expression ${\cos }^{-1}\left[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right]$ is $\underset{\_}{168°}$.

(b)

To determine

The value of ${\sin }^{-1}\left[\frac{\overrightarrow{A}\times \overrightarrow{B}}{AB}\right]$.

Answer to Problem 6P

The value of the given expression ${\sin }^{-1}\left[\frac{\overrightarrow{A}\times \overrightarrow{B}}{AB}\right]$ is $\underset{\_}{11.9°}$

Explanation of Solution

Write the expression for cross product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ as.

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|\\ =\left|\begin{array}{cc}{A}_y& A_z\\ B_y& B_z\end{array}\right|\widehat{i}+\left|\begin{array}{cc}{A}_z& A_x\\ B_z& B_x\end{array}\right|\widehat{j}+\left|\begin{array}{cc}{A}_x& A_y\\ B_x& B_y\end{array}\right|\widehat{k}\end{array}$

Simplify the above-obtained expression as.

  $\overrightarrow{A}\times \overrightarrow{B}=\left(A_y{B}_z-A_z{B}_y\right)\widehat{i}+\left(A_z{B}_x-A_x{B}_z\right)\widehat{j}+\left(A_x{B}_y-A_y{B}_x\right)\widehat{k}$                      (V)

Write the expression for the magnitude of $\overrightarrow{A}\times \overrightarrow{B}$ the vector as.

  $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left({\left(AB\right)}_x\right)}^2+{\left({\left(AB\right)}_y\right)}^2+{\left({\left(AB\right)}_z\right)}^2}$                                              (VI)

Here, ${\left(AB\right)}_x$ is $x$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$, ${\left(AB\right)}_y$ is $y$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$, ${\left(AB\right)}_z$ is $z$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$.

Write the expression for the angle between the vectors.

  $\theta ={\sin }^{-1}\left[\frac{\overrightarrow{A}\times \overrightarrow{B}}{AB}\right]$

Substitute $\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}$ for $A$, $\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}$ for $B$ and $\sqrt{{\left({\left(AB\right)}_x\right)}^2+{\left({\left(AB\right)}_y\right)}^2+{\left({\left(AB\right)}_z\right)}^2}$ for $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$ in given expression as.

  $\theta ={\sin }^{-1}\left[\frac{\left(\sqrt{{\left({\left(AB\right)}_x\right)}^2+{\left({\left(AB\right)}_y\right)}^2+{\left({\left(AB\right)}_z\right)}^2}\right)}{\left(\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}\right)\left(\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}\right)}\right]$             (VII)

Conclusion:

Substitute $-3$ for $A_x$, $7$ for $A_y$, $-4$ for $A_z$, $-2$ for $B_x$, $-10$ for $B_y$ and $9$ for $B_z$ in equation (I).

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left(\left(-3\right)\left(9\right)-\left(-4\right)\left(-10\right)\right)\widehat{i}+\left(\left(-3\right)\left(9\right)-\left(-4\right)\left(6\right)\right)\widehat{j}+\left(\left(-10\right)\left(-3\right)-\left(6\right)\left(7\right)\right)\widehat{k}\\ =23.00\widehat{i}+3.00\widehat{j}-12.00\widehat{k}\end{array}$

Substitute $23.00$ for ${\left(AB\right)}_x$, $3.00$ for ${\left(AB\right)}_y$, $-12$ for ${\left(AB\right)}_z$, $-3$ for, $A_x$, $7$ for $A_y$, $-4$ for $A_z$, $6$ for $B_x$, $-10$ for $B_y$ and $9$ for $B_z$ in equation (VII).

  $\begin{array}{c}\theta ={\sin }^{-1}\left[\frac{\sqrt{{\left(23.00\right)}^2+{\left(3.00\right)}^2+{\left(-12.00\right)}^2}}{\left(\sqrt{{\left(-3\right)}^2+{\left(7\right)}^2+{\left(-4\right)}^2}\right)\left(\sqrt{{\left(6\right)}^2+{\left(-10\right)}^2+{\left(9\right)}^2}\right)}\right]\\ ={\sin }^{-1}\left[\frac{\sqrt{682}}{\left(\sqrt{74}\right)\left(\sqrt{217}\right)}\right]\\ =11.88°\\ \approx 11.9°\end{array}$

Thus, the value of the given expression ${\sin }^{-1}\left[\frac{\overrightarrow{A}\times \overrightarrow{B}}{AB}\right]$ is $\underset{\_}{11.9°}$.

(c)

To determine

The correct method which gives the angle between two vectors.

Answer to Problem 6P

The first method gives the angle between the two vectors more accurately.

Explanation of Solution

Write the expression for the angle between the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ as.

  $\theta ={\cos }^{-1}\left[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right]$                                                                                   (VIII)

Conclusion:

The RHS expression of the equation (VIII) is similar to expression asked in first part.

The angle calculation between two vectors will be correct when using the first method.

Generally, the angle between the two vectors can only be at most $180°$. If use the second method for calculation angle, than it can’t distinguish $\theta$ from $\left(180-\theta \right)$ because $\sin \left(180-\theta \right)=\sin \theta$.

Thus, the first method gives the angle between the two vectors more accurately.